Question

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 40 manufacturing companies. Estimate the mean, median, and standard deviation of advertising expense.
Advertising Expenditure ($millions) Number of companies

$20 to under $30 9

30 to under 40 13

40 to under 50 21

50 to under 60 18

60 to under 70 14

Total 75

Answer 1

Answer:

Mean = 47

Median = 47.38

Standard Deviation = 12.73

Explanation:

Note: You wrote " 40 manufacturing companies, but the total number of companies you actually listed is 75, definitely you meant 75.

Let y represent the range of advertising expenditure, f represent the number of companies, x represent the midpoint of the range of advertising expenditure.

y                                       f                      x                  fx                    fx²

$20 to under $30            9                     25               225               5625

$30 to under $40            13                    35               455               15925

$40 to under $50            21                    45               945              42525

$50 to under $60            18                    55               990              54450

$60 to under $70            14                     65               910               59150

                                       n = 75                           $\sum fx = 3525$      

$\sum fx^2 = 177675$

Mean, $\bar{X} = \frac{\sum fx}{n}$

$\bar{X} = \frac{3525}{75} \\\bar{X} = 47$

Standard Deviation:

$SD = \sqrt{\frac{n \sum fx^2 - (\sum fx)^2}{n(n-1)} } \\SD = \sqrt{\frac{(75*177675) - (3525)^2}{75(75-1)} }\\SD = 12.73$

Median:

Get the cumulative frequencies(cf)

         y                                        f                                cf

$20 to under $30                     9                                9

$30 to under $40                     13                               22

$40 to under $50                     21                               43

$50 to under $60                     18                               61

$60 to under $70                      14                              75

                                                N = 75

Median = Size of (N/2)th item

Median = Size of (75/2)th item

Median = Size of (37.5)th item

The median class = 40 to under 50

Lower limit, L₁ = 40

Cumulative frequency, cf = 22

f = 21

Class Width, h = 10

Median = $L_1 + \frac{ (N/2) - cf}{f} * h$

Median = $40 + \frac{ (75/2) - 22}{21} * 10$

Median = 47.38